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let $a,b,c\ge 0$,and such $abc=1$,show that

$$a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$$

My solution: Without loss of generality,assume that

$a=\max{(a,b,c)}$, since $abc=1$,we have $a\ge 1$,

we will show that $$f(a,b,c)\ge f(a,t,t)\ge 0, t=\sqrt{bc},0<t\le 1$$

since $$f(a,b,c)-f(a,t,t)=(\sqrt{b}-\sqrt{c})^2[(\sqrt{b}+\sqrt{c})^2+8a-10]$$ then equivalent to $$(\sqrt{b}+\sqrt{c})^2+8a\ge 10$$ which is true because $$(\sqrt{b}+\sqrt{c})^2+8a\ge 4\sqrt{bc}+8a=4(a+\sqrt{bc})+4a\ge 8\sqrt{a\sqrt{bc}}+4a=8\sqrt[4]{a}+4a\ge 12$$

Now,since $a=\dfrac{1}{t^2}$,we have $$f(a,t,t)=f(\dfrac{1}{t^2},t,t)=\dfrac{(10t^4-7t^2+2t+1)(t-1)^2}{t^4}$$ which is clearly nonnegative because

$$10t^4-7t^2+2t+1=(3t^2-1)^2+t(1-t)+t^4+t>0$$

Have other nice methods? Thank you

math110
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2 Answers2

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Let $f(x) = x^2+\dfrac8x + 1 - 10x + 16 \log x$ for $x > 0$.
Then the given inequality is $f(a)+f(b)+f(c) \ge 0$, and it is sufficient to show $f(x) \ge 0$.

We note $f'(x) = \dfrac{2(x-2)^2 (x-1)}{x^2}$.
Thus for $x < 1, f'(x)<0$ and for $x > 1, f'(x) \ge 0$. Hence $\forall x >0, \; f(x) \ge f(1) = 0$.

Macavity
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    where does the 16 lnx came from thank you – Will Mar 29 '22 at 08:04
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    @GregoriusWillson When facing the constraint $abc=1$, a good way to separate the variables is by taking logs. Thus adding $k\cdot \log x$ to $f(x)$ will not change anything, as on summing across $a, b, c$, it will disappear. What is left is to choose $k$ wisely, so that $f(x)\geqslant 0$ in the domain of interest. In this case we know equality is when $a=b=c=1$, so we choose $k$ s.t. $f'(1)=0$ and test for non-negativity... – Macavity Mar 29 '22 at 09:31
  • thanks im just realizing it nice solution btw – Will Mar 29 '22 at 10:21
  • but how if the roots of first derivative are not double rooted like for this case this is $(x-2)^2*(x-1)$ how if $\frac{(2(x-1)(x^2-29x+55))}{x^2}$ – Will Mar 29 '22 at 11:58
  • There is no general prescription here, in many cases this approach would lead to a simple proof - but not in all cases. Maybe a more general approach (for 'separable' inequalities) would be to try Jensen / Karamata first, then tangent line or this one, then more complex approaches like substitutions or mixing variables if earlier approaches fail... – Macavity Mar 29 '22 at 13:36
  • do u have email or any private chat because i would ask u something about problem i had – Will Mar 29 '22 at 13:41
  • @GregoriusWillson Suggest you post separately if there's a different problem, i or someone else should help out. – Macavity Mar 29 '22 at 13:51
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Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality is a linear inequality of $v^2$, which says that

it's remains to prove the last inequality for an extremal value of $v^2$,

which happens for equality case of two variables.

Let $b=a$ and $c=\frac{1}{a^2}$.

Hence, we need to prove that $$(a-1)^2(10a^4-7a^2+2a+1)\geq0,$$ which is obvious.

Michael Rozenberg
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